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0-pi-e-2x-sin-x-dx-

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Question Number 77918 by john santu last updated on 12/Jan/20
∫ _0 ^π e^(−2x) sin x dx ?
∫π0e−2xsinxdx?
Commented by mathmax by abdo last updated on 12/Jan/20
∫_0 ^π e^(−2x) sinx dx =Im(∫_0 ^π e^(−2x+ix) dx) we have ∫_0 ^π e^((−2+i)x) dx =[(1/(−2+i))e^((−2+i)x) ]_0 ^π =(1/(−2+i))(e^((−2+i)π) −1) =((−1)/(2−i)){ −e^(−2π) −1) =(1/(2−i))(e^(−2π) +1) =((2+i)/5)(e^(−2π) +1) =(2/5)(1+e^(−2π) )+(1/5)(e^(−2π) +1)i ⇒ ∫_0 ^π e^(−2x) sinxdx =((1+e^(−2π) )/5)
∫0πe−2xsinxdx=Im(∫0πe−2x+ixdx)wehave∫0πe(−2+i)xdx=[1−2+ie(−2+i)x]0π=1−2+i(e(−2+i)π−1)=−12−i{−e−2π−1)=12−i(e−2π+1)=2+i5(e−2π+1)=25(1+e−2π)+15(e−2π+1)i⇒∫0πe−2xsinxdx=1+e−2π5
Answered by john santu last updated on 12/Jan/20

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